<span>The gain in potential energy is 600*2 = 1200 J, and this is the minimum required work that would be required in a frictionless situation. The work that actually has to be done is F*X = 300*5 = 1500 J. The efficiency is therefore 1200/1500 = 80%</span>
B. Because you can't perform an experiment that is impossible to perform
The pipe that produces the highest-frequency sound is the 5 cm long pipe.
According to the Fundamental Principle, the length of the tube of the pipe and its frequency is inversely proportional.
This means that the longer the tube, the lower the frequency and vice versa. Therefore, between the 25cm long pipe and the 5cm long pipe, the shorter pipe produces the highest-frequency sound.
We first calculate the work effectively done by the force when it compressed the steel cylinder. This is solved by multiplying the distance that the cylinder was compressed by and its causing force.
0.010 m * 13,000 N = 130 N-m
Then, we solve for the work done by the 1,500 N force:
0.10 m * 1,500 N = 150 N-m
We then take the ratio of the efficient:experimental
130 N-m / 150 N-m = 0.87 = 87%
Answer:
70 N
21°
1.1 m/s²
Explanation:
Draw a free body diagram of the block. There are three forces:
Weight pulling straight down
Normal force pushing perpendicular to the incline
Friction force pushing parallel to the incline
Part 1
Sum the forces in the perpendicular direction:
∑F = ma
N − mg cos θ = 0
N = mg cos θ
The block is at rest, so F = N μs:
F = N μs
F = mg μs cos θ
F = (20 kg) (9.8 m/s²) (0.38) (cos 19°)
F = 70 N
Part 2
Sum the forces in the parallel direction (down the incline is positive):
∑F = ma
mg sin θ − F = 0
mg sin θ = N μs
mg sin θ = mg μs cos θ
tan θ = μs
θ = atan μs
θ = atan 0.38
θ = 21°
Part 3
Sum the forces in the parallel direction (this time, acceleration is not 0).
∑F = ma
mg sin θ − F = ma
mg sin θ − N μk = ma
mg sin θ − mg μk cos θ = ma
a = g (sin θ − μk cos θ)
a = (9.8 m/s²) (sin 24° − 0.32 cos 24°)
a = 1.1 m/s²
